Statement-1 : Rotational KE of a gas molecule follows Maxwell's speed distribution curve.
Statement-2: Rotational KE & translational KE of a diatomic gas molecule is same,

Options

A. 1-true 2-false
B. 1-false 2-true
C. 1-false 2-false
D. 1-true 2-false

Answer

To solve the given question, we will analyze both statements step by step.

### Step 1: Analyze Statement 1
**Statement 1**: "Rotational KE of a gas molecule follows Maxwell's speed distribution curve."

- The Maxwell speed distribution describes the distribution of speeds of particles in an ideal gas at a given temperature.
- It states that at a specific temperature, gas molecules have a range of speeds, and the distribution can be represented graphically.
- The rotational kinetic energy (KE) of gas molecules, particularly diatomic molecules, can be related to their speed. Since the speed distribution is governed by Maxwell's distribution, it implies that the rotational KE also follows this distribution.

**Conclusion for Statement 1**: This statement is **true**.

### Step 2: Analyze Statement 2
**Statement 2**: "Rotational KE & translational KE of a diatomic gas molecule is the same."

- For a diatomic gas molecule, the degrees of freedom (DOF) are important. A diatomic molecule has:
- 3 translational degrees of freedom (movement in x, y, z directions).
- 2 rotational degrees of freedom (rotation about two axes perpendicular to the line connecting the two atoms).
- The translational kinetic energy (TKE) can be expressed as:
\[ TKE = \frac{3}{2} kT \]
where \( k \) is the Boltzmann constant and \( T \) is the temperature.
- The rotational kinetic energy (RKE) can be expressed as:
\[ RKE = \frac{2}{2} kT = kT \]
- Comparing the two:
- Translational KE = \( \frac{3}{2} kT \)
- Rotational KE = \( kT \)

**Conclusion for Statement 2**: Since \( \frac{3}{2} kT \neq kT \), this statement is **false**.

### Final Conclusion
- Statement 1 is true.
- Statement 2 is false.

Thus, the correct answer is that Statement 1 is true and Statement 2 is false.